---
title: "Fourier series: x − x²"
subject: "Mathematics III"
chapter: 4
key: "semester_3/mathematics_iii:fourier-series-x-x2-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/fourier-series-x-x2-two
site: BCT Engineering
---

# Fourier series: x − x²

Mathematics III · Chapter 4: Fourier Series

- Asked in: 2073 Chaitra, 5 marks; 2070 Asar, 5 marks (asked 2 times)
- Video: [Fourier Series of (x-x^2): Full Calculation](https://www.youtube.com/watch?v=yMWKgKi0qDk) — Ayan Sarkar
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/fourier-series-x-x2-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-4

## Question

Find the Fourier series to represent f(x) = x − x² from −π to π.

## Short answer

f(x)=x-x^2 is neither even nor odd. a_0 = -2pi^2/3, a_n = 4(-1)^(n+1)/n^2, b_n = 2(-1)^(n+1)/n. The Fourier series is x-x^2 = -pi^2/3 + sum_{n=1}^infinity [4(-1)^(n+1)/n^2 cos(nx) + 2(-1)^(n+1)/n sin(nx)].

## Answer

f(x) = x − x² on (−π, π). Use ∫ x² cos nx dx and ∫ x sin nx dx over (−π, π), with cos nπ = (−1)ⁿ.

#### Coefficients
```text
a₀ = (1/π) ∫_{−π}^{π} (x − x²) dx = (1/π)(−2π³/3) = −2π²/3
aₙ = (1/π) ∫ (x − x²) cos nx dx = −(1/π)∫ x² cos nx dx = −4(−1)ⁿ/n²
bₙ = (1/π) ∫ (x − x²) sin nx dx = (1/π)∫ x sin nx dx = 2(−1)ⁿ⁺¹/n
```

(x cos nx and x² sin nx are odd, so they integrate to 0; ∫_{−π}^{π} x² cos nx dx = 4π(−1)ⁿ/n² and ∫_{−π}^{π} x sin nx dx = 2π(−1)ⁿ⁺¹/n.)

#### Series
```text
x − x² = −π²/3 + Σ_{n=1}^∞ [ 4(−1)ⁿ⁺¹/n² · cos nx + 2(−1)ⁿ⁺¹/n · sin nx ]
       = −π²/3 + 4cos x − cos 2x + (4/9)cos 3x − ...
         + 2 sin x − sin 2x + (2/3) sin 3x − ...
```

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which part of x - x^2 gives the sine series?
   A. x^2
   B. both
   C. neither
   D. x
   Answer: D. x is odd. (Key point: Split)

2. a0 =
   A. -2 pi^2/3
   B. 2 pi^2/3
   C. -pi^2/3
   D. pi
   Answer: A. (1/pi)(-2 pi^3/3). (Key point: a0)

3. an =
   A. 4(-1)^n/n^2
   B. 4(-1)^(n+1)/n^2
   C. 2(-1)^(n+1)/n
   D. 0
   Answer: B. Equals -4(-1)^n/n^2. (Key point: an)

4. bn =
   A. 4(-1)^(n+1)/n^2
   B. 2(-1)^n/n
   C. 2(-1)^(n+1)/n
   D. 0
   Answer: C. From integral of x sin nx. (Key point: bn)

## Close variants set in other exams

- [Fourier series: 2x − x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-2x-x2-three) — Mathematics III, chapter 4, asked 2×

## Related in Mathematics III

- [Fourier series: x − x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x-x2) — Mathematics III, chapter 4, asked 6×
- [Fourier series of a function](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-of-a-function) — Mathematics III, chapter 4, asked 6×
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- [Fourier series: x − x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x-x2-four) — Mathematics III, chapter 4
- [Fourier series: x²](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x2) — Mathematics III, chapter 4, asked 9×
- [Fourier series: x sin x](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Afourier-series-x-sin-x-two) — Mathematics III, chapter 4, asked 7×

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